When trigonometry meets complexity

Our friend Kenny Lau has stated on his "extension" the method I used. I think it is nice to show how it is done. So here it goes: the Maclaurin series for the sine function calculated at $i$ is

$\sin(i) = i \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)!} i^{2n} = i \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)!} (-1)^{n} = i \sum_{n=0}^{\infty} \frac{1}{(2n+1)!}$

But

$\sum_{n=0}^{\infty} \frac{1}{(2n+1)!} = \sinh(1) = \frac{e - e^{-1}}{2} = \frac{e^{2} - 1}{2e}$

Then

$\sin(i) = i \sinh(1) = \frac{i}{2e} (e^{2} - 1)$

So

$a + b = \sinh(1) \approx 1.175$

Euler's formula : $e^{ix}=\cos x+i\sin x$

Substituting $x$ as $i$ gives: $e^{i\times i}=\cos i+i\sin i$

Using the definition of $i$ : $e^{-1}=\cos i+i\sin i\ \ -(1)$

Substituting $x$ as $-i$ gives: $e^{i\times-i}=\cos(-i)+i\sin(-i)$

Using the definition of $i$ and using some basic trigonometric identities: $e^{1}=\cos i-i\sin i\ \ -(2)$

Subtracting $(2)$ from $(1)$ : $e^{-1}-e^1=2i\sin i$

Therefore: $\sin i=\frac1{2i}\left(\frac1e-e\right)=\left(\frac e2-\frac1{2e}\right)$

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Appendix:

Definition of $i$ :

• $i^2=-1$ (taking only one root)

Trigonometric identities used:

• $\cos(-\theta)=+\cos\theta$
• $\sin(-\theta)=-\sin\theta$

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Extension: $\sin(ix)=i\sinh(x)=\left(\frac{e^x-e^{-x}}2\right)i$ (The proof is left as an exercise for the reader.)